Let $({f_n})_{n\geq 1}$ be a sequence of functions on $[0,1]$ such that $\lim
_{n\rightarrow \infty}f_{n}(x)=f(x)$ almost everywhere with respect to Lebesgue measure and
$$\sup_{n\geq 1}||f_n||_{L^4}<\infty.$$

Prove that
$$\lim_{n\rightarrow \infty}||f_n - f||_{L^3} = 0.$$

Since we're on a finite measure space I see how we get that $f_n \in L^3$. But I don't see how I can use a convergence theorem on $L^p$ larger than one.

3 Answers
3

A sequence bounded in $L_p$, $p>1$, is uniformly integrable. Using this fact with $p=4/3$, it follows that $(f_n^3)$ is uniformly integrable. It now follows from Vitali's Convergence Theorem that $f_n$ converges to $f$ in the $L_3$-norm.

Since $f_n$ converges pointwise, for any $\epsilon>0$ Egorov's Theorem states that except on $E_{\large\epsilon}$, a set of measure $\epsilon$, $f_n\to f$ uniformly. Therefore,
$$
\lim_{n\to\infty}\int_{\large E_{\Large\epsilon}^c}\left|f_n(x)-f(x)\right|^3\,\mathrm{d}x=0\tag{2}
$$
Furthermore, by Hölder's Inequality with $p=4/3$ and $q=4$, we have
$$
\limsup_{n\to\infty}\int_{\large E_{\Large\epsilon}}\left|f_n(x)-f(x)\right|^3\,\mathrm{d}x\le\epsilon^{1/4}\sup_n\|f_n-f\|_{L^4}^3\tag{3}
$$
Since $\epsilon$ is arbitrary, we get
$$
\lim_{n\to\infty}\int_0^1\left|f_n(x)-f(x)\right|^3\,\mathrm{d}x=0\tag{4}
$$